Problem: $ \left(\dfrac{1}{64}\right)^{-\frac{4}{3}}$
Answer: $= 64^{\frac{4}{3}}$ $= \left(64^{\frac{1}{3}}\right)^{4}$ To simplify $64^{\frac{1}{3}}$ , figure out what goes in the blank: $\left(? \right)^{3}=64$ To simplify $64^{\frac{1}{3}}$ , figure out what goes in the blank: $\left({4}\right)^{3}=64$ so $ 64^{\frac{1}{3}}=4$ So $64^{\frac{4}{3}}=\left(64^{\frac{1}{3}}\right)^{4}=4^{4}$ $= 4\cdot4\cdot 4\cdot 4$ $= 16\cdot4\cdot 4$ $= 64\cdot4$ $= 256$